Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MX = 100;
int arr[MX][MX];
int Min[2][MX];
int n, m;
int cnt;
bool x;
void Swap(int n, int m) {
  if (n > m) {
    int c = m;
    m = n;
    n = c;
  }
}
void Scan() {
  if (n > m) {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) cin >> arr[j][i];
    Swap(n, m);
  } else
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) cin >> arr[i][j];
}
void Value() {
  for (int i = 0; i < n; i++) Min[0][i] = 501;
  for (int j = 0; j < m; j++) Min[1][j] = 501;
}
void Find_Min_R() {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) Min[0][i] = min(Min[0][i], arr[i][j]);
    cnt += Min[0][i];
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) arr[i][j] -= Min[0][i];
}
void Find_Min_C() {
  for (int j = 0; j < m; j++) {
    for (int i = 0; i < n; i++) Min[1][j] = min(Min[1][j], arr[i][j]);
    cnt += Min[1][j];
  }
}
bool Check() {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
      if (arr[i][j] > Min[1][j]) return false;
  return true;
}
void Print() {
  if (n > m) {
    int c = m;
    m = n;
    n = c;
  }
  cout << cnt << endl;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < Min[0][i]; j++)
      if (x == true) {
        cout << "col ";
        cout << i + 1 << endl;
      } else
        cout << "row " << i + 1 << endl;
  for (int j = 0; j < m; j++)
    for (int i = 0; i < Min[1][j]; i++)
      if (x == true) {
        cout << "row ";
        cout << j + 1 << endl;
      } else
        cout << "col " << j + 1 << endl;
}
int main() {
  cin >> n >> m;
  Scan();
  if (n > m) {
    int c = m;
    m = n;
    n = c;
    x = true;
  }
  Value();
  Find_Min_R();
  Find_Min_C();
  if (Check() == true)
    Print();
  else
    return cout << -1, 0;
}